- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 4. Common Probability Distributions
- Subject 6. Normal Distribution
CFA Practice Question
For x, a normal random variable, which of the following is (are) false?
II. P(x < 38) = P(x < 38)
III. The higher the standard deviation the more spread out the distribution.
IV. P(35 < x < 50) = P(x <50) - P(x < 35)
I. P(x < mean) = 50%
II. P(x < 38) = P(x < 38)
III. The higher the standard deviation the more spread out the distribution.
IV. P(35 < x < 50) = P(x <50) - P(x < 35)
Correct Answer: I
P(x < mean) = 50%. A normal distribution is centered (symmetrical about) the mean. The area under the curve to the left of the mean, P(x <= mean), is 0.5. The area under the curve to the right of the mean, P(x >= mean), is 0.5. The total area under the bell-shaped curve is 1.
User Contributed Comments 18
| User | Comment |
|---|---|
| tony1973 | I believe the answer tries to emphasize that x can be equal to the mean, therefore P(x < mean) is actually less than 50%. You can have a normal distribution in which some observations' values are exactly the mean, therefore P(x<mean) must be less than 50% if it is a normal distribution. |
| mordja | If P(x<= mean) = P(x<mean) + p(x=mean) If we learnt in the previous question that the probability of any actual value equals zero, then p(x=mean)=0 Therefore it follows that P(x<=mean) = p(x<mean) Thus p(x<mean) must equal p(x>mean), therefore p(x<mean) + p(x>mean) = 1 (as the area under the distribution is 1), thus p(x<mean) = 1/2....... |
| jmcohen87 | i agree |
| DannyZhou | Am I insane or are you guying not reading the question. (I) is correct. But the question is asking "which of the following is/are false?". So the correct answer is none of them. |
| Poorvi | DannyZhou - the answer is that I is false. This is because P(x<=mean) is 0.5 and not P(x<mean) is 0.5 |
| kenchew1 | I is false because a normal distribution can be skewed and have kurotisis. If it is skewed, P(x<median) = 50%, not P(x<mean). |
| LionHero | I is true since for any single point on a continuous distribution, the probability is 0; thus P(X<mean) = P(X<=mean) |
| gazza77 | Kenchew1, a normal distribution cannot be skewed and has no excess kutosis |
| gazza77 | is a nornal distribution also a continuous distribution? If so, say for example p(x=4)=0 therefore p(x=mean)=0 therefore P(x<mean) = P(x<=mean) therefore I is true and all the choices are false |
| jmumm | I, too, agree with the mordja's logic. |
| tll936 | is that mean the ans. false? |
| eddington | Analyst Notes people, help on this one...... |
| CFAIcand15 | any further insight on this one? |
| nfressell2 | It may be rounded up to 50% but since it cannot contain the mean it is less than 50%. I.e it is 49.99999999999999999% |
| Fabulous1 | MordJa, your logic is funny. If you'd follow it properly then since all the single values are zero, also the sum of those values would be zero. Even if a single value is zero it still has to be included in the calc of the appropriate integral to solve for the prob of an interval. |
| Akiva | Fabulous, if the single value is zero, it does not mean that the sum of them will be zero. Mordja, jmcohen87, LionHero, Gazza77, +1. I is true. |
| myron | Akiva: if the single value is zero, it does not mean that the sum of them will be zero. What did you really want to say? |
| EMMADADA | I agreed with DannyZhou. All of them are correct. So which of the following is(are) false? The correct answer should be, none of them. |